# Proving an isometric dilation of a non unitary operator on Hilbert space implies infinite dimensional space involving matrices

I have been given this exercise in my Operator theory class dealing with operators on Hilbert spaces, which reads as follows:

Let H be a Hilbert space. We are to prove, in two distinct ways, that if $T \in B(H)$ is a contraction (Contraction: An operator T satisfying $||T|| \leq 1$) which is not unitary, and if V is an isometric dilation of T, then the larger Hilbert space K, satisfying $H \subset K$, is necessarily infinite dimensional.

Clarification: By isometric dilation I simply mean that the extended operator is itself an isometry.

Edit: the context for this is the Sz.-Nagy's dilation theorem as seen here https://en.wikipedia.org/wiki/Sz.-Nagy%27s_dilation_theorem And also from the text book by Nagy: Here is an arxiv link to this result on page 1 the Nagy theorem: http://arxiv.org/pdf/1012.4514.pdf

I unfortunately have to say that I truly have no idea on this one even after working hard at this, I don't have one way to prove this claim even in one method I thought assuming to get contradiction that I can assume K is finite dimensional and then work with matrices but I got nothing. I am stuck and truly desperate.

• $V$ is an isometric dilation of $T$ ? $V = \alpha T$ with $|\alpha| > 1$ ? and I don't understand what is $K$, and how it is related to $T$ and $V$ – reuns Mar 20 '16 at 17:59
• @user1952009 : isometric dilation as in the sense of the Sz.-Nagy dilation theorem – kroner Mar 20 '16 at 18:03
• @user1952009 : the dilation is to some isometry on a larger Hilbert space K in this sense – kroner Mar 20 '16 at 18:05
• wait, $H \subset K$ hence $K \simeq H \bigoplus H^\perp$, $T : H \to H$, $V : K \to K$, $\|T\| \le 1$ and there is $x$ such that $\|Tx\| < \|x\|$ ($T$ is not unitary), and for every $y \in K$ : $\|Vy\| = \| y\|$ ($V$ is unitary) and for every $x \in H$ : $Tx = P Vx$ where $P : K \to H$ is the orthogonal projection onto $H$ i.e.for $x \in H, u \in H^\perp$ : $P(x+u) = x$. that's it ? – reuns Mar 20 '16 at 18:13
• what if $H$ is finite dimensional (hence $T\sum_{i=1}^n a_i e_i = \sum_{i=1}^n a_i T e_i$ ) ? how to we prove that $K$ is infinite dimensional ? – reuns Mar 20 '16 at 18:19

## 1 Answer

If $T \in \mathcal{B}(H)$ is a contraction, then $T$ has a unitary dilation on a Hilbert space $K$, meaning that there exists an isometry $V : H\rightarrow K$ and a unitary $U$ on $H$ such that $T^{n} = V^{\star}U^{n} V$ for all $n \ge 0$. Automatically $(T^{\star})^{n}=V^{\star}(U^{\star})^{n}V$.

Proof 1: Suppose $K$ is finite dimensional. Then $U$ is isometric and, hence, injective, which makes $U$ surjective; so $U$ must be unitary in the case of finite-dimensional $K$. If $m$ is the minimal polynomial for $U$, then $m(0) \ne 0$, which leads to a polynomial $q$ such that $q(U)=U^{-1}$. $T$ is normal because $$T^{\star} = V^{\star}U^{\star}V = V^{\star}U^{-1}V=V^{\star}q(U)V = q(T)$$ And $T$ must be unitary because \begin{align} T^{\star}T & = (V^{\star}U^{\star}V)(V^{\star}UV) \\ & =(V^{\star}q(U)V)(V^{\star}UV) \\ & = V^{\star}q(U)UV= V^{\star}V = I. \end{align}

Proof 2: It appears you found a second proof in the comments.

• I thank you very much for a nicely written answer – kroner Mar 31 '16 at 11:00
• @zbigniew2015 : You're welcome. It was interesting looking into this topic again. Good question. – DisintegratingByParts Mar 31 '16 at 11:07